Mathematical Induction 4+11+14+21+....+(5n+(-1)^n

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The discussion focuses on proving the formula for the sum of the series 4 + 11 + 14 + 21 + ... + (5n + (-1)^n) using mathematical induction. The user successfully verifies the base case for n=1 and establishes the induction hypothesis. They outline the steps taken to prove the formula for k+1, including algebraic manipulations to incorporate the new term. The proof concludes by demonstrating that the derived expression for P_{n+1} aligns with the expected format, thus completing the induction process. The user seeks assistance with the algebraic steps but ultimately arrives at a valid proof.
Yankel
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Dear all

I am trying to prove by induction the following:

View attachment 8712

I checked it for n=1, it is valid. Then I assume it is correct for some k, and wish to prove it for k+1, got stuck with the algebra. Can you kindly assist ?

Thank you.
 

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I would state the induction hypothesis \(P_n\):

$$\sum_{k=1}^{n}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5n(n+1)+(-1)^n-1\right)$$

As the induction step, I would add $$5(n+1)+(-1)^{n+1}$$ to both sides:

$$\sum_{k=1}^{n}\left(5k+(-1)^k\right)+5(n+1)+(-1)^{n+1}=\frac{1}{2}\left(5n(n+1)+(-1)^n-1\right)+5(n+1)+(-1)^{n+1}$$

Incorporate the new term:

$$\sum_{k=1}^{n+1}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5n(n+1)+(-1)^n-1+2\cdot5(n+1)+2\cdot(-1)^{n+1}\right)$$

$$\sum_{k=1}^{n+1}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5(n+1)(n+2)+(-1)^n(1+2(-1))-1\right)$$

$$\sum_{k=1}^{n+1}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5(n+1)((n+1)+1)+(-1)^{n+1}-1\right)$$

We have derived \(P_{n+1}\) from \(P_n\) thereby completing the proof by induction.
 
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